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**Bladescope** I seen a question on a practice paper a while back which confused me a 'lil. Here it is:

Source: [6663]Edexcel GCE Core Mathematics C1 AS - Monday 10th Januaryy 2005

Question 9

The gradient of the curve C is given by

$\displaystyle \frac{dy}{dx}

=(3x-1)^2$

The point P(1,4) lies on C. (For reference, but was used for questions A and B)

c) Using dy/dx, show that there is no point on C at which the tangent is parallel to the line

$\displaystyle y=1-2x$

I don't know where to start on this question, so any help would be great :x, especially since I have a C1 exam in ~ 2 hours. D:

So far, all I know is that the gradient of $\displaystyle y=1-2x$ is obviously -2 and the increment (+1) doesn't matter (hence, parallel).

$\displaystyle (3x-1)^2=9x^2-6x+1$

So those are the 2 gradients of the equasion, and I have no idea on how to show that there's no point on C where they're parallel.